Respuesta :

[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=\frac{1}{3}\\ n=5\\ a_5=\frac{1}{3} \end{cases}\implies a_5=a_1\cdot \left( \frac{1}{3} \right)^{5-1}[/tex]

[tex]\bf \cfrac{1}{3}=a_1\cdot \left( \frac{1}{3} \right)^{5-1}\implies \cfrac{1}{3}=a_1\cdot \cfrac{1^4}{3^4}\implies \cfrac{1}{3}=\cfrac{a_1}{81}\implies \cfrac{81}{3}=a_1 \\\\\\ 27=a_1\qquad \qquad thus\qquad \qquad \boxed{a_n=27 \left( \frac{1}{3} \right)^{n-1}}[/tex]